If â and b̂ are two unit vectors and θ is the angle between them prove that sin θ 2 = 1 2 |â − b̂|.

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The problem statement, all variables and given/known data given two unit vectors a= cosθi + sinθi b=cosΦi+sinΦj prove that sin(θ-Φ)=sinθcosΦ-cosΦsinθ using vector algebra
Relevant equations sin(θ-Φ)=sinθcosΦ-cosΦsinθ
The attempt at a solution axb= (cosθsinΦ-cosΦsinθ)k and I'm guessing that the change in sign has something to do with the fact that k is perpendicular to the vectors I'm usingΦ-θ

when I calculated the components I got axb = ( cosθi +sinθj )x(cosΦi + sinΦj) axb=cosθcosΦixi + cosθsinΦixj +sinθcosΦjxi +sinθsinΦjxj ixi=1x1xsino=0 jxjxsin0=0 ixj=1x1sin90=1 and jxi=-1 because AxB=-BxA and was then left with axb= cosθsinΦ(1) + sinθcosΦ(-1)=cosθsinΦ -sinθcosΦ but its supposed to be the other way around, I dont understand where I'm going wrong :(

I was trying this problem but I can't solve with the $\sin(\theta /2)$.

I try this:

$$<\mathbf{u}+\mathbf{v},\mathbf{u}>\ = \|\mathbf{u}+\mathbf{v} \| \|\mathbf{u} \| \cos\alpha$$ where $\alpha$ is the angle between $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}$.

$$ <\mathbf{u}+\mathbf{v},\mathbf{v}>\ = \| \mathbf{u}+\mathbf{v} \| \|\mathbf{v}\| \cos\alpha$$

Then I add the equations:

$$<\mathbf{u}+\mathbf{v},\mathbf{u}+\mathbf{v}>\ = 2 \| \mathbf{u}+\mathbf{v} \| \cos\frac{\theta}{2}$$

Because $\| \mathbf{u}\| = 1$ and $\| \mathbf{v}\| = 1$. So,

$$\frac{\| \mathbf{u}+\mathbf{v} \|}{2} = \cos \frac{\theta}{2}$$ Maybe I forgot some explanations, but that is my idea however, I can't see how to go from $\cos$ to $\sin$.

Maybe there is another form; thanks for your help in advance!

Last updated at Jan. 23, 2022 by